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Solution. Find the side M of a square equivalent to the figure P, and the side N of a square equivalent to the figure Q. Then let X be a fourth proportional to the three given lines M, N, AB; upon the side X, homologous to AB, describe a figure similar to the figure P; it will be equivalent to the figure Q. For, by calling Y the figure described upon the side X, we shall have

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We have also, by construction, MP, and N=Q;

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hence YQ; therefore the figure Y is similar to the figure P and equivalent to the figure Q.


257. To construct a rectangle equivalent to a given square C (fig. 152), and whose adjacent sides shall make a given sum AB. Solution. On AB, as a diameter, describe a semicircle, and draw DE parallel to the diameter, and at a distance AD, equal to a side of the given square C. From the point E, in which the parallel cuts the circumference, let fall upon the diameter the perpendicular EF; AF and FB will be the sides of the rectangle sought.

For their sum is equal to AB, and their rectangle AF × FB is equal to the square of EF (215), or of AD; therefore the rectangle is equivalent to the given square C.

258. Scholium. It is necessary in order that the problem may be possible, that the distance AD should not exceed the radius, that is, that the side of the square should not exceed half of the line AB.

Fig. 152.


259. To construct a rectangle equivalent to a square C (fig. 153), Fig. 153. and whose adjacent sides shall differ by a given quantity AB.

Fig. 154.

Solution. On the given line AB, as a diameter, describe a circle; from the extremity of the diameter draw the tangent AD equal to the side of the square C. Through the point D and the centre O draw the secant DF; DE and DF will be the adjacent sides of the rectangle required.

For, 1. the difference of the sides is equal to the diameter EF or AB; 2. the rectangle DE × DF is equal to AD (228); therefore this rectangle will be equivalent to the given square C.


260. To find the common measure, if there be one, between the diagonal and side of a square.

Solution. Let ABCG (fig. 154) be any square, and AC its diagonal.

We are, in the first place, to apply CB to CA, as often as it can be done (157); and in order to this, let there be described, from the centre C, and with a radius CB, the semicircle DBE. It will be seen, that CB is contained once in AC with a remainder AD; the result of the first operation therefore is the quotient 1 with the remainder AD, which is to be compared with BC, or its equal AB.

We may take AF=AD, and apply AF actually to AB; and we should find that it is contained twice with a remainder. But, as this remainder and the following ones go on diminishing and would soon become too small to be perceived, on account of the imperfection of the mechanical operation, we can conclude nothing with regard to the question, whether the lines AC, CB, have a common measure or not. Now there is a very simple method, by which we may avoid these decreasing lines, and which only requires an operation to be performed upon lines of the same magnitude.

The angle ABC being a right angle, AB is a tangent, and AE is a secant, drawn from the same point, so that

AD: AB:: AB: AE (228).

Thus, in the second operation which has for its object to compare AD with AB, we may, instead of the ratio of AD to AB, take that of AB to AE. Now AB, or its equal CD, is contained twice in AE with a remainder AD; therefore the result of the second operation is the quotient 2 with the remainder AD, which is to be compared with AB.

The third operation, which consists in comparing AD with AB, reduces itself likewise to comparing AB, or its equal CD, with AE, and we have still the quotient 2 with the remainder AD.

Whence it is evident that the operation will never terminate, and that accordingly there is no common measure between the diagonal and the side of a square, a truth already made known by a numerical operation, since these two lines are to each other :: 2:1 (188), but which is rendered clearer by the geometrical solution.

261. It is not then possible to find in numbers the exact ratio of the diagonal to the side of a square; but we may approximate it to any degree we please by means of the continued fraction which is equal to this ratio. The first operation gave for a quotient 1; the second and each of the others continued without end gives 2; thus the fraction under consideration becomes

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If, for example, we take the four first terms of this fraction, we find that its value added to the first quotient 1 is 113, or 41; so that the approximate ratio of the diagonal to the side of a square is :: 41: 29. The ratio might be found more exactly by taking a greater number of terms.


Of regular Polygons and the Measure of the Circle.


262. A POLYGON, which is at the same time equiangular and equilateral, is called a regular polygon.

Regular polygons admit of any number of sides. The equilateral triangle is one of three sides; and the square one of four.


263. Two regular polygons of the same number of sides are similar figures.

Demonstration. Let there be, for example, the two regular hexagons ABCDEF, abcdef (fig. 155); the sum of the angles is Fig. 155.

†See note on continued fractions at the end of Lacroix's Arithmetic.

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the same in both, and is equal to eight right angles (82). The angle A is the sixth part of this sum as well as the angle a; therefore the two angles A and a are equal; the same may be said of the angles B and b, C and c, &c.

Moreover, since by the nature of these polygons the sides AB, BC, CD, &c., are equal, as also a b, bc, cd, &c.,

AB: ab:: BC: bc:: CD: cd, &c.;

consequently the two figures under consideration have their angles equal and their homologous sides proportional; therefore they are similar (162).

264. Corollary. The perimeters of two regular polygons of the same number of sides are to each other as their homologous sides, and their surfaces are as the squares of these sides (221).

265. Scholium. The angle of a regular polygon is determined by the number of its sides like the angle of an equiangular polygon (79).

Fig. 156.


266. Every regular polygon may be inscribed in a circle and may be circumscribed about a circle.

Demonstration. Let ABCDE &c. (fig. 156), be any regular polygon, and let there be described a circle, whose circumference shall pass through the three points A, B, C; let O be its centre, and OP a perpendicular let fall upon the middle of the side BC; join AO and OD.

The quadrilateral OPCD may be placed upon the quadrilateral OPBA; in fact the side OP is common, and the angle OPC=OPB, each being a right angle, consequently the side PC will fall upon its equal PB, and the point Cupon B. Moreover, by the nature of the polygon, the angle PCD=PBA; therefore CD will take the direction BA, and CD being equal to BA, the point D will fall upon A, and the two quadrilaterals will coincide throughout. Hence the distance OD is equal to the distance OA, and the circumference, which passes through the three points A, B, C, will pass also through the point D. similar reasoning it may be shown, that the circumference, which passes through the three vertices B, C, D, will pass through the next vertex E, and so on; therefore the same circumference, which passes through the three points A, B, C,


passes through all the vertices of the angles of the polygon, and the polygon is inscribed in this circumference.

Furthermore, with respect to this circumference, all the sides AB, BC, CD, &c. are equal chords; they are accordingly equally distant from the centre (109); if therefore from the point O, as a centre, and with the radius OP, a circle be described, the circumference will touch the side BC and all the other sides of the polygon, each at its middle point, and the circle will be inscribed in the polygon, or the polygon will be circumscribed about the circle.

267. Scholium 1. The point O, the common centre of the inscribed and circumscribed circle, may be regarded also as the centre of the polygon; and for this reason we call the angle of the centre the angle AOB formed by the two radii drawn to the extremities of the same side AB.

Since all the chords AB, BC, &c., are equal, it is evident that all the angles at the centre are equal, and that the value of each is found by dividing four right angles by the number of the sides of the polygon.

268. Scholium II. In order to inscribe a regular polygon of a certain number of sides in a given circle, it is only necessary to divide the circumference into as many equal parts as the polygon has sides; for, the arcs being equal, the chords AB, BC, CD, &c. (fig. 158), will be equal; the triangles ABO, BOC, Fig. 158. COD, &c., will also be equal, for the sides of the one will be respectively equal to those of the other; consequently all the angles ABC, BCD, CDE, &c., will be equal; therefore the figure ABCDE &c. will be a regular polygon.


269. To inscribe a square in a given circle.

Solution. Draw the diameters AC, BD (fig. 157), cutting Fig. 157. each other at right angles; join the extremities A, B, C, D, and

the figure ABCD will be the inscribed square.

For, the angles AOB, BOC, &c., being equal, the chords AB, BC, &c. are equal.

270. Scholium. The triangle BOC being right-angled and isosceles, BC: BO :: √2:1 (188); therefore, the side of an inscribed square is to radius, as the square root of two is to unity.



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